Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $x = \dfrac{-10a + 30}{a - 1} \div \dfrac{10a + 10}{a^2 - 1} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{-10a + 30}{a - 1} \times \dfrac{a^2 - 1}{10a + 10} $ First factor the quadratic. $x = \dfrac{-10a + 30}{a - 1} \times \dfrac{(a - 1)(a + 1)}{10a + 10} $ Then factor out any other terms. $x = \dfrac{-10(a - 3)}{a - 1} \times \dfrac{(a - 1)(a + 1)}{10(a + 1)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ -10(a - 3) \times (a - 1)(a + 1) } { (a - 1) \times 10(a + 1) } $ $x = \dfrac{ -10(a - 3)(a - 1)(a + 1)}{ 10(a - 1)(a + 1)} $ Notice that $(a + 1)$ and $(a - 1)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -10(a - 3)\cancel{(a - 1)}(a + 1)}{ 10\cancel{(a - 1)}(a + 1)} $ We are dividing by $a - 1$ , so $a - 1 \neq 0$ Therefore, $a \neq 1$ $x = \dfrac{ -10(a - 3)\cancel{(a - 1)}\cancel{(a + 1)}}{ 10\cancel{(a - 1)}\cancel{(a + 1)}} $ We are dividing by $a + 1$ , so $a + 1 \neq 0$ Therefore, $a \neq -1$ $x = \dfrac{-10(a - 3)}{10} $ $x = -(a - 3) ; \space a \neq 1 ; \space a \neq -1 $